Question
The dimensions of $$\frac{1}{2}{\varepsilon _0}{E^2},$$ where $${\varepsilon _0}$$ is permittivity of free space and $$E$$ is electric field, are
A.
$$\left[ {M{L^2}{T^{ - 2}}} \right]$$
B.
$$\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$
C.
$$\left[ {M{L^2}{T^{ - 1}}} \right]$$
D.
$$\left[ {ML{T^{ - 1}}} \right]$$
Answer :
$$\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$
Solution :
As we know that,
Dimension of $${\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$$
Dimension of $$E = \left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]$$
So, dimension of $$\frac{1}{2}{\varepsilon _0}{E^2} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \times {\left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]^2}$$
$$ = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$$